Question

If hydrochloric acid is obtained commercially at a concentration of 12.1M, how many milliliters of 12.1M HCl(aq) must be used to prepare 2.00x103mL of 0.500M HCL(aq)?

Answer 1

`V_1=82.6mL`

Step-by-step explanation:

Hello there!

In this case, according to this question, we will need to deal with this dilution problem, because it is asking for the volume of a 12.1-M stock solution of HCl. In such a way, we can use the following equation, under the assumption of no change in the number of moles in the solution:

`M_2V_2=M_1V_1`

Thus, we solve for the initial volume, V1, as shown below:

`V_1=(M_2V_2)/(M_1)`

And plug in the initial concentration and final concentration and volume to obtain:

`V_1=(2000mL*0.500M)/(12.1M)\\\\V_1=82.6mL`

Regards!